Question

eqn1. KIO3 + 5KI + 3H2S04 --> 3I2 +3H2O

eqn2. C6H8O6 + I2 --> C6H606 + 2HI

eqn3. 2Na2S2O3 + I2 --> 2NaI +Na2S4O6

Stock KIO3 solution = .02453mol/L

a) 25mL of stock KIO3 is diluted with deionized water to 250mL, what is the concentration of the dilute KIO3 solution ?

b) 25mL of the dilute solution is pipetted into a flask, 0.2g of KI is added, and 1 mL of 1M sulfuric acid, how many moles of I2 is produced from the 25mL of dilute KIO3?

Answer 1

a) Given Initially stock, [KIO₃] = 0.02453 mol/L = 0.02453 M.

We need to know final [KIO₃] = ? on diluting 25 mL of initial stock KIO₃ solution to 250.0 mL.

Let, Initially, M1 = 0.02453 M, V1 = 25.0 mL

Finally, M2 = ? M, V2 = 250.0 mL

By Law of dilution,

M2V2 = M1V1

M2 x 250.0 = 0.02453 x 25.0 mL

M2 = 0.02453 x 25.0 / 250.0

M2 = 0.002453 M.

b) Let us calculate first mmol of KIO3 in 25.0 mL of this diluted solution.

mmol of KIO3 = Molarity of KIO3 x volume in mL = 0.002453 x 25.0 = 0.06133 mmol.

Also calculate mmol of KI in 0.2 g

Molar mass of KI = 166.003 g/mol

# moles of KI = Given mass / Molar mass = 0.2 / 166.003 = 0.0012 moles

mmol of KI = 0.0012 moles = 0.0012 x 1000 = 1.2 mmol of KI.

Eq.1 : KIO₃ + 5KI + 3H₂SO₄ --> 3I₂ +3H₂O

mmol relationship can be given as

1 mmol of KIO₃ 5 mmol of KI 3 mmol of I₂.

mmol of KIO₃ = 0.06133 mmol ....... (calculated above)

mmol of KI = 1.2

It's obvious that KI is in very much large excess and hence KIO₃ is a limiting reagent which governs the yield of product.

From mole relationship above,

If, 1 mmol of KIO₃ 3 mmol of I₂.

Then, 0.06133 mmol of KIO₃ = 3 x 0.06133 = 0.184 mmol of I₂.

0.184 mmol = 0.184 x 10^-3 mol = 1.84 x 10^-4 mol

Assuming 100 completion of reaction,

Answer b) : 1.84 x 10^-4 mol of I₂ will be produced from 25.0 mL of diluted stock KIO3 solution.

===========================XXXXXXXXXXXXXXXXXXXXXX=======================